Problem: In triangle $ABC$, $AB = 13$, $AC = 15$, and $BC = 14$.  Let $I$ be the incenter.  The incircle of triangle $ABC$ touches sides $BC$, $AC$, and $AB$ at $D$, $E$, and $F$, respectively.  Find the length of $BI$.
Answer: Since $AE$ and $AF$ are tangents from the same point to the same circle, $AE = AF$.  Let $x = AE = AF$.  Similarly, let $y = BD = BF$ and $z = CD = CE$.

[asy]
import geometry;

unitsize(2 cm);

pair A, B, C, D, E, F, I;

A = (1,2);
B = (0,0);
C = (3,0);
I = incenter(A,B,C);
D = (I + reflect(B,C)*(I))/2;
E = (I + reflect(C,A)*(I))/2;
F = (I + reflect(A,B)*(I))/2;

draw(A--B--C--cycle);
draw(incircle(A,B,C));

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NE);
label("$F$", F, NW);
label("$x$", (A + E)/2, NE);
label("$x$", (A + F)/2, NW);
label("$y$", (B + F)/2, NW);
label("$y$", (B + D)/2, S);
label("$z$", (C + D)/2, S);
label("$z$", (C + E)/2, NE);
[/asy]

Then $x + y = AB = 13$, $x + z = AC = 15$, and $y + z = BC = 14$.  Adding all these equations, we get $2x + 2y + 2z = 42$, so $x + y + z = 21$.  Subtracting the equation $x + z = 15$, we get $y = 6$.

By Heron's formula, the area of triangle $ABC$ is \[K = \sqrt{21(21 - 14)(21 - 15)(21 - 13)} = 84,\]so the inradius is $r = K/s = 84/21 = 4$.

Hence, by Pythagoras on right triangle $BDI$, \[BI = \sqrt{BD^2 + DI^2} = \sqrt{y^2 + r^2} = \sqrt{6^2 + 4^2} = \sqrt{52} = \boxed{2 \sqrt{13}}.\]